Factor $f(x)$ into linear factors given that $k$ is a zero of $f(x)$.
$f(x)=2 x^3-3 x^2-5 x+6 ; k=1$
A. $f(x)=(x-1)(x-2)(2 x+3)$
B. $f(x)=(x-1)(x+1)(2 x-6)$
C. $f(x)=(x-1)(x+2)(2 x-3)$
D. $f(x)=(x+1)(x+2)(2 x-3)$

To factor the polynomial f ( x ) = 2 x 3 − 3 x 2 − 5 x + 6 , we first identify that ( x − 1 ) is a factor because k = 1 is a zero. The complete factorization is f ( x ) = ( x − 1 ) ( x − 2 ) ( 2 x + 3 ) , which corresponds to option A.
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Answered by Anonymous | 2025-07-04

Recognize that since k = 1 is a zero, ( x − 1 ) is a factor.
Divide the polynomial by ( x − 1 ) to get the quadratic factor 2 x 2 − x − 6 .
Factor the quadratic 2 x 2 − x − 6 into ( x − 2 ) ( 2 x + 3 ) .
Write the complete factorization: f ( x ) = ( x − 1 ) ( x − 2 ) ( 2 x + 3 ) .

f ( x ) = ( x − 1 ) ( x − 2 ) ( 2 x + 3 ) ​
Explanation

Problem Analysis We are given the cubic polynomial f ( x ) = 2 x 3 − 3 x 2 − 5 x + 6 and told that k = 1 is a zero of f ( x ) . This means that f ( 1 ) = 0 , and thus ( x − 1 ) is a factor of f ( x ) . Our goal is to factor f ( x ) into linear factors.

Polynomial Division Since we know that ( x − 1 ) is a factor, we can perform polynomial division to find the other factor, which will be a quadratic. Dividing 2 x 3 − 3 x 2 − 5 x + 6 by ( x − 1 ) gives us 2 x 2 − x − 6 .

Factoring the Quadratic Now we need to factor the quadratic 2 x 2 − x − 6 . We are looking for two numbers that multiply to 2 ( − 6 ) = − 12 and add up to − 1 . These numbers are − 4 and 3 . So we can rewrite the quadratic as: 2 x 2 − x − 6 = 2 x 2 − 4 x + 3 x − 6

Factoring by Grouping Now we factor by grouping: 2 x 2 − 4 x + 3 x − 6 = 2 x ( x − 2 ) + 3 ( x − 2 ) = ( 2 x + 3 ) ( x − 2 )

Complete Factorization Therefore, the complete factorization of f ( x ) is: f ( x ) = ( x − 1 ) ( x − 2 ) ( 2 x + 3 )

Final Answer Comparing this with the given options, we see that the correct factorization is ( x − 1 ) ( x − 2 ) ( 2 x + 3 ) .


Examples
Factoring polynomials is a fundamental concept in algebra and has many real-world applications. For example, engineers use polynomial factorization to analyze the stability of structures and systems. Imagine designing a bridge; the load distribution can be modeled by a polynomial equation. By factoring this polynomial, engineers can identify critical points where the structure might be weak or unstable, ensuring the bridge's safety and longevity. Similarly, in control systems, factoring polynomials helps in determining the stability of feedback loops, crucial for designing efficient and reliable systems.

Answered by GinnyAnswer | 2025-07-04